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kobusy [5.1K]
3 years ago
8

A student is told that both energy and mass must be conserved in every chemical reaction. She measures the mass of Hydrochloric

acid and a zinc strip separately. She then places the zinc strip into the acid and bubbles form as the zinc looks like it disappears. The combined mass afterward is less than the original.
Chemistry
2 answers:
salantis [7]3 years ago
6 0

Answer:

Explanation:

When we react Hydrochlorid Acid with zinc we have the following reaction:

2HCl(aq) + Zn(s) --> ZnCl2(aq) + H2(g)

The hydrogen gas formed is lost to the environment, so we can affirme that in the start we have the mass for all the H, Cl and Zn atoms in the solution, but after the reaction occurs, we have only the mass for the Cl and Zn atoms.

That's why the mass is less than the original.

The law that the student was told is only applied to closed environments.

Eva8 [605]3 years ago
4 0
I think the combined mass was less afterward than the original mass due to escape of hydrogen gas in the atmosphere. According to the law of conservation of mass, mass is conserved and does not change in a chemical reaction which occurs in a closed system. However in an open system the mass may not be conserved and may change due to release of gases to the atmosphere. Like in this case, hydrogen gas produced in the reaction was released to the atmosphere thus reducing the final mass.
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Would you classify hydrogen as a metal or a non-metal? <br>Explain why.​
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Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H
lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

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